Area of triangle with a perpendicular problem.
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Given that triangle PRQ is a right angle. A line is perpendicular to PQ at point N on the line and it meets at R.
$NR=2.4cm$ and $QR=3cm$
How do I find the area of the triangle PRQ?
My teacher said I can't used trigonometry
$PR$ is the base.
$Area=frac{bh}{2}=frac{3PR}{2}$
area
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up vote
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favorite
Given that triangle PRQ is a right angle. A line is perpendicular to PQ at point N on the line and it meets at R.
$NR=2.4cm$ and $QR=3cm$
How do I find the area of the triangle PRQ?
My teacher said I can't used trigonometry
$PR$ is the base.
$Area=frac{bh}{2}=frac{3PR}{2}$
area
Here is a link to what the triangle looked like google.co.uk/…:
– time
Nov 21 at 16:54
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given that triangle PRQ is a right angle. A line is perpendicular to PQ at point N on the line and it meets at R.
$NR=2.4cm$ and $QR=3cm$
How do I find the area of the triangle PRQ?
My teacher said I can't used trigonometry
$PR$ is the base.
$Area=frac{bh}{2}=frac{3PR}{2}$
area
Given that triangle PRQ is a right angle. A line is perpendicular to PQ at point N on the line and it meets at R.
$NR=2.4cm$ and $QR=3cm$
How do I find the area of the triangle PRQ?
My teacher said I can't used trigonometry
$PR$ is the base.
$Area=frac{bh}{2}=frac{3PR}{2}$
area
area
asked Nov 21 at 16:50
time
6
6
Here is a link to what the triangle looked like google.co.uk/…:
– time
Nov 21 at 16:54
add a comment |
Here is a link to what the triangle looked like google.co.uk/…:
– time
Nov 21 at 16:54
Here is a link to what the triangle looked like google.co.uk/…:
– time
Nov 21 at 16:54
Here is a link to what the triangle looked like google.co.uk/…:
– time
Nov 21 at 16:54
add a comment |
2 Answers
2
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1
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First find $NQ=sqrt{RQ^2-NR^2}$. Then use similarity of triangles $NRQ$ and $PRQ$: $frac{NQ}{RQ}=frac{RQ}{PQ}$ to find $PQ$.
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You can find it using the trigonometric ratios sine and cosine. This picture shows the working more clearly.

Once you have PR, you can find area- Area= (QR)*(PR)/2 = 6.
1
No trigs allowed ...
– Michael Hoppe
Nov 21 at 19:33
Wow, I've got to read questions better. Sorry!
– karun mathews
Nov 23 at 2:47
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
First find $NQ=sqrt{RQ^2-NR^2}$. Then use similarity of triangles $NRQ$ and $PRQ$: $frac{NQ}{RQ}=frac{RQ}{PQ}$ to find $PQ$.
add a comment |
up vote
1
down vote
First find $NQ=sqrt{RQ^2-NR^2}$. Then use similarity of triangles $NRQ$ and $PRQ$: $frac{NQ}{RQ}=frac{RQ}{PQ}$ to find $PQ$.
add a comment |
up vote
1
down vote
up vote
1
down vote
First find $NQ=sqrt{RQ^2-NR^2}$. Then use similarity of triangles $NRQ$ and $PRQ$: $frac{NQ}{RQ}=frac{RQ}{PQ}$ to find $PQ$.
First find $NQ=sqrt{RQ^2-NR^2}$. Then use similarity of triangles $NRQ$ and $PRQ$: $frac{NQ}{RQ}=frac{RQ}{PQ}$ to find $PQ$.
answered Nov 21 at 17:11
Vasya
3,3091515
3,3091515
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add a comment |
up vote
0
down vote
You can find it using the trigonometric ratios sine and cosine. This picture shows the working more clearly.

Once you have PR, you can find area- Area= (QR)*(PR)/2 = 6.
1
No trigs allowed ...
– Michael Hoppe
Nov 21 at 19:33
Wow, I've got to read questions better. Sorry!
– karun mathews
Nov 23 at 2:47
add a comment |
up vote
0
down vote
You can find it using the trigonometric ratios sine and cosine. This picture shows the working more clearly.

Once you have PR, you can find area- Area= (QR)*(PR)/2 = 6.
1
No trigs allowed ...
– Michael Hoppe
Nov 21 at 19:33
Wow, I've got to read questions better. Sorry!
– karun mathews
Nov 23 at 2:47
add a comment |
up vote
0
down vote
up vote
0
down vote
You can find it using the trigonometric ratios sine and cosine. This picture shows the working more clearly.

Once you have PR, you can find area- Area= (QR)*(PR)/2 = 6.
You can find it using the trigonometric ratios sine and cosine. This picture shows the working more clearly.

Once you have PR, you can find area- Area= (QR)*(PR)/2 = 6.
answered Nov 21 at 17:06
karun mathews
223
223
1
No trigs allowed ...
– Michael Hoppe
Nov 21 at 19:33
Wow, I've got to read questions better. Sorry!
– karun mathews
Nov 23 at 2:47
add a comment |
1
No trigs allowed ...
– Michael Hoppe
Nov 21 at 19:33
Wow, I've got to read questions better. Sorry!
– karun mathews
Nov 23 at 2:47
1
1
No trigs allowed ...
– Michael Hoppe
Nov 21 at 19:33
No trigs allowed ...
– Michael Hoppe
Nov 21 at 19:33
Wow, I've got to read questions better. Sorry!
– karun mathews
Nov 23 at 2:47
Wow, I've got to read questions better. Sorry!
– karun mathews
Nov 23 at 2:47
add a comment |
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Here is a link to what the triangle looked like google.co.uk/…:
– time
Nov 21 at 16:54